Verified Solution • 2025–26 Session moderate

QJEEM22Phee45: JEE Main 2022 Physics - Class 12 JEEM22Phee45 Solution: A plane electromagnetic wave travels in a medium of relative…

A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 ×\times 10 −- 2 Am −- 1 at a point, what will be the approximate magnitude of electric field intensity at that point?

(Given : Permeability of free space μ\mu 0 = 4 π\pi ×\times 10 −- 7 NA −- 2, speed of light in vacuum c = 3 ×\times 108 ms −- 1)

Multiple Choice Options

  • A 16.96 Vm −- 1
  • B 2.25 ×\times 10 −- 2 Vm −- 1
  • C 8.48 Vm −- 1
  • D 6.75 ×\times 106 Vm −- 1

Step-by-Step Verified Solution

H = 4.5 ×\times 10 −- 2

So B = μ\mu 0 μ\mu H

Thus E=cnBE = {c \over n}B (where n ⇒\Rightarrow refractive index)

So E=3×108×4π×10−7×1.61×4.5×10−21.61×6.44E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}

E=8.48E = 8.48

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